External pressure on welded steel pipe 2

[sbw]

I'm trying to determine the maximum allowable external pressure on a 54"x0.375" and a 54x0.5" welded steel pipe (grade A36). Specifically, the pipe is capped at the bottom standing in water, so I'm trying to determine maximum allowable depths for the external hydraulic head. Ordinarily I'd order the relevent design manuals, which in this case I believe to be AWWA M11 and AISI vol 3, but I've been asked to respond by tomorrow a.m.

Can someone direct me to on-line design criteria or software that I can use in this pinch? (I've already used the 30-day 'free' trial on the AISI software).

 

[Cockroach]

You can apply the Von Mises-Hencky Model for triaxial stress on a wall element. This is perhaps the best way to model it, like a thick wall pressure vessel subjected to external, not bore pressure.

Noting your material, ASTM A36 I would get 573 psig for the 54 OD X 3/8 inch wall and 763 psig for the 54 OD X 1/2 inch walled pipe, respectively.

Noting water is approximately 62.5 lbf/ft^3, this would correspond to 1320 ft in case A and 1758 ft in case B, respectively.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[CRG]

Look at this thread:

hollow cylinders under high external pressure
thread794-95854

 

[warrenw]

Cockroach;
Are you saying that a 54" diam. pipe with a 3/8" wall can support an external pressure of 573 psi without buckling?
I find that difficult to believe.
You might want to check your numbers.
Warren

 

[Cockroach]

Yeah, that is what I'm saying.

Buckling implies a load axial through the centre line, I would prefer collapse since there is no indication that the pipe is a load bearing member. This is an external pressure, circumferential to the OD and would more properly represent the minimal pressure to take that specified pipe to yield, i.e. FS=1.

573 psig on a 54 OD pipe of 3/8 wall would experience a stress of 36 ksi, i.e. material yield.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[Cockroach]

WarrenW:

Even if you improperly modeled the pipe as a thin wall pressure vessel, using S=PD/2t would give:

S = [(573 psig)(54.0 in)]/[2(0.375 in)] = 41.3 ksi

The material yield is 36 ksi, so we're there.

In reality, you can't use this simple of an analysis because of the wall thickness. That equation only applies given D/t<10 and we're at 54.0/0.375 = 144.

I stand by my model and calculations.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[sbw]

Cockroach,

If I'm not mistaken the formula for internal pressure is based on the radius.

As an aside, I was able to get a liscensed copy of the software that I had made reference to late this morning (turns out the 2-3day delay promissed in the original message was off by 2-3 days). Based on this program, the calculated maximum external buckling pressures are on the order of 17 psi for the the 0.375" wall and 40 psi for the 0.500" wall. My main objection with these types of programs is that they function as a black box, which doesn't help with the actual understanding of how the critical buckling capacity is calculated.

If you're still interested, there's more to the problem. The application is a vertical liner for a shaft about 70' deep. A 6' hole will be drilled in rock to depth, the casing set and restrained and the void grouted solid. I've been told that the pipe should be checked for head pressure with water at the ground surface. So, these are my questions:
1. does it really make sense to design for that kind of uniform pressure when contact between casing (including grout) and water will limited to the fractures in the rock?
2. can I add the capacity of the grout for a composite section with steel lining and grout?

They've got the 0.375" pipe sitting next to the hole, so they'd really like to see it work, if justifiable.

Thanks for the input.

 

[CRG]

I believe Cockroach’s approach assumes that the pipe is a perfect cylinder with walls that are of a consistent thickness, not the 12.5% mill tolerance that is common. If you make those assumptions he is correct; however, life is never so convenient. The reality is that the pipe is both imperfect in roundness and thickness. As such the, “Theory of Elastic Stability,” by S. Timoshenko is of relevant concern. Hence the difference between Coackroach’s calculation and AISI software is more than just a safety factor, it reflects the imperfect construction of pipe.

 

[Cockroach]

Actually, you're right, that simple equation is based on radius. That's why there is a "2" in the denominator. At any rate, you can see what I am saying regarding collapse pressure ONLY. You may have other factors here.

I don't quite believe the 17 psig buckling pressure for 54 OD X 3/8 inch ASTM A36 pipe. You need a lot more pressure to collapse the tube, she's got 36,000 psi yield! This is a very common practice in the oilfield, casing is subjected to tremendous pressures during cementing without collapse. It is true that casing is typically not 54 OD, but none-the-less, the calculation is fully supported by experience. Your application sounds very similar, except grout fills the annulus to the open hole rather than cement. I'm talking 1200 kg/m^3 cement here, so at 1.1 km down you can see the head!

You may need to reshuffle the equation to account for weight of tubing above the bottom. This would change the nature of your question, an axial load is beyond the scope of my calculation since I've considered orientation to be in the horizontal. Also, loading may be ramped, atmospheric at the top, but at formation pressure so many feet below the surface. This too is not modeled.

I have done the calculation in the past, based on biaxial stress with longitudinal loading and obtained very close results to strain gauge laboratory results. By far the Von Mises-Hencky model is superior to other methods. Perhaps I will fool around with the model over a few beer and see what becomes of it.


Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[sbw]

Thanks for the additional input. I should have been more clear in my previous statement that the pressures are allowable and are based on 50% of the yield, or 18 ksi, per the program guidelines. For what it's worth, I ran it for 12" ODx0.375" pipe can arrived at an allowable external pressure of 1600 psi.

In terms of the stress in pipe from internal pressures, the equation is S=pr/2t, per Timoshenko 'Mechanics of Materials'.

Thanks again.

 

[jte]

sbw-

The stress in the pipe that you are referring to is the longitudinal stress due to external pressure. The hoop stress would be S=PR/t, or twice the long. stress. In any case, it is essentially meaningless to evaluate the stress in a situation where column buckling or collapse of the pipe is concerned. The programs you have are giving similar results to a Section VIII external pressure calc. I'd suggest that you check your design per Section VIII. The trick you have is that you have an external pressure which varies along the length of the pipe. The easy and conservative way out is to assume that the entire shell is subject to the max. head.

jt

 

[prex]

sbw,
look thread404-91966 for a formula for buckling of thin walled cylinders. For an infinite length cylinder (like yours) this formula reduces to p_cr=0.25(E/(1-[&nu;]²))(t³/R³) that, for your 54"ODx0.375" pipe would give a critical external pressure of 1.5 bar (22 psi?), to be reduced with a suitable safety factor (usually 3) to obtain an allowable external pressure. So the results you obtain with your software are realistic, and even a bit optimistic.
However all the above is for an unrestrained pipe subject to external pressure. What comes into consideration is not really the grout making a composite section with the steel, as I suppose they are unconnected, but the fact that the grout restrains the pipe from going outwards. Now the buckling phenomenon represented by the formulae requires not only the wall to go inwards, as one would expect, but also part of it to go outwards, or buckling will not occur.
So you are correct in asking whether it is realistic to take the allowable pressure of an unrestrained cylinder.
However this doesn't close the problem, as local buckling (where a small portion of the wall bumps inwards) may still take place and might be activated by imperfections in the grout filling or by cracks.
I'm afraid that here no formula can be provided, and only experience with similar constructions can help. I also guess that at this point you would require a code to support your choice, but I have none to suggest.
Concerning Cockroach's approach, I suppose that his figures are for relatively thick cylinders, where the primary cause of instability is plastic collapse, so would be very careful in taking them as a basis for evaluation.

prex

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